* Reading Assignments:

1.1 1.1.1 1.1.21.2 1.2.1 1.2.2

2.1 2.1.1 2.1.2 2.1.3 2.1.4

2. Basic Concepts of Thermodynamics

2.1 Thermodynamic system A specified collection of matter is called a system, which is defined by the mass and the composition. a. Open system: mass is exchanged with its surroundings;

b. Closed system: NO mass is exchanged with its surroundings.

What type of system does atmospheric thermodynamics deal with?

The systems that atmospheric thermodynamics deal with include

1) an air parcel; 2) a cloud;3) the atmosphere;4) an air mass etc.

Precisely speaking, they are open systems because mass canbe changed by the entrainment and mixing processes.

But, we will treat them as a closed system in this course.

Assumptions:

1) the volume is large that mixing at the edges is negligible; or

2) the system is imbedded in a much larger mass which has the same properties.

2.2 Thermodynamic properties

The properties define the thermodynamic state of a system.

a. Intensive property: does not depend on the mass (m) or does not change with subdivision of the system, denoted by lowercase letters, e.g., z.

b. Extensive property: does depend on the mass (m) or does change with subdivision of the system, denoted by uppercase letters, e.g., Z.

Exception to the convention: T for temperature and m for massException to the convention: T for temperature and m for mass

* An intensive property is also called a specific property if

For example, volume V is an extensive property, so v=V/m (i.e., volume per unit mass) is a specific property and an intensive property.

m

Zz

a. A system is considered to be homogeneous if every intensive property has the same value for every point of the system.

b. A system is said to be heterogeneous if the intensive property of one portion is different from the property of another portion.

mzZ

ii

i zmZ

* Homogeneous vs heterogeneous

* A system can exchange energy with its surroundings through two mechanisms:

1) Mechanical exchange (Expansion work)

performing work on the surroundings

2) Thermal exchange (Heat transfer)

transferring heat across the boundary

* A system is in thermodynamic equilibrium if it is in mechanical and thermal equilibrium.

Mechanical equilibrium: the pressure difference between the system and its surroundings is infinitesimal;

Thermal equilibrium: the temperature difference between the system and its surroundings is infinitesimal.

2.3 Expansion work

If a system is not in mechanical equilibrium with its surrounding it will expand or contract.

The incremental expansion work:

p: the pressure exerted by the surroundings over the system

dV: the incremental volume

dS: the displaced section of surface

dn: the normal distance between original and expanded surface

p

pdVpdSdnW

2.4 Heat transfer

Adiabatic process: no heat is exchanged between the system and the environment.

Diabatic process: heat is exchanged between the system and the environment.

Which one will we use the most? Why?

2.5 State variables and equation of state

* A system, if its thermodynamic state is uniquely determined by any two intensive properties, is defined as a pure substance. The two properties are referred to as state variables.

* From any two state variables, a third can be determined by an equation of state,

.0),,( 321 zzzf

A pure substance only has two degrees of freedom. Any two state variables fix the thermodynamic state,

* Any third state variable as a function of the two independent state variables forms a state surface of the thermodynamic states, i.e.,

).,( 213 zzgz

2.6 Thermodynamic process

* The transformation of a system between two states describes a path, which is called a thermodynamic process.

* There are infinite paths to connect two states.

* Exact differentials

Consider

If

.),(),( dyyxNdxyxMz

we have

,dzdyy

zdxx

zz

is an exact differential, is a point functionwhich is path independent,

,),(,),(y

zyxN

x

zyxM

dz ),( yxz

which is the same as

,),(),(

x

yxN

y

yxM

.0dz

2.7 Equation of state for ideal gases

2.7.1 How to obtain the ideal gas equation?

The most common way to deduce fundamental equations is toobserve controlled experiments.

* Based on Boyle’s observation, if the temperature of a fixed mass of gas is constant, the volume of the gas (V) is inversely proportional to its pressure (p), i.e.,

constpV

* From Charles’ observation, for a fixed mass of gas at constant pressure, the volume of the gas is directly proportional to its absolute temperature (T), i.e.,

constT

V

(1)

(2)

* For a fixed mass of gas, consider three different equilibrium states that have , respectively.),,(),,,(),,,( 00 TVpandTVpTVp ssss

* From (1) and (2), we have

0

0,T

V

T

VVppV ssss

0

0

T

Vp

T

pV ss

Combine them,

Divide (3) by the molar abundance (or number of moles)

which is constant since the mass (m) and molecular weight (M) are constant, we have

(4)

(3)

M

mn

nT

Vp

Tn

pV ss

0

0

* For a standard condition,

11

0

0* 3143.8 KJmolnT

VpR ss

is called the universal gas constant.

Now, (4) can be rearranged to get the equation of state for the ideal gas

TnRpV * (5)

molmnV

skgmmbatmpKT

s

s

/104.22/

1001325.125.10131,15.27333

0

2150

2.7.2 Equivalent forms of ideal gas equation

Ideal gas equation (5) can be written in several forms,

mRTTRM

mpV * (6)

M

RR

*

is the specific gas constant.

Since the specific volume

1

m

Vv

(6) can be also written as

, is the density,

RTporRTpv , (7)

2.7.3 Equation of state for mixture of ideal gases

Each gas obeys its own state equation, for the th gasi

Since in a mixture of gases,

* The partial pressure is: the pressure the th gas would have if the same mass existed alone at the same temperature and occupied the same volume as the mixture;

* The partial volume is: the volume the th gas would occupy if the same mass existed alone at the same temperature and pressure.

i

i

ip

iV

iiiii TRmVp (8)

(8) can be written in form,

TRmVp iii Sum (9) over all gases in the mixture, and apply Dalton’s law,

(9)

i

ipp

we get the equation of state for the mixture,

TRmpV (10)

is the mean specific gas constantm

RmR i

ii

i

imm

which is similar to the ideal gas equation (6).

The mean molecular weight of the mixture is defined by

n

m

n

m

n

MnM i

ii

ii

Since

i i

i

ii M

mnn

(11)

(11) can be written as

R

R

Rm

mR

MRm

mRM

iii

iii

**

*

*

)/(

The molar fraction is used to measure the relative concentrationof the th gas over the total abundance air in the mixture,i

n

nN ii

Using the state equations for the th gas and the mixture of gases, wecan also have

i

V

V

p

p

n

nN iiii

The mass fraction is also used to measure the relative concentration.

Using iii MmnandMmn /,/ in (12), we can get

M

MN

M

M

n

n

m

m ii

iii

(12)

The absolute concentration of the th gas is measured by its density . ii

The mixing ratio is used to measure the relative concentration of the th gas over dry air, e.g., the mass mixing ratio is defined in form,i

d

ii m

mr (13)

is the mass of dry air; is dimensionless and expressed in for tropospheric water vapor.dm ir

1kgg

We can also have the volume mixing ratio related to the molar fraction,

i

i

i

i

d

i

N

N

VV

V

V

V

1

Since the mass of air in the presence of water vapor and ozone is virtuallyidentical to the mass of dry air, (13) can be related to the molar fraction,

iid

i

d

i

dd

ii

d

ii N

M

M

n

n

n

n

Mn

Mn

m

mr (14)

where 1, dd

ii n

n

M

M

(15)

(d)

2.8 Atmospheric composition

Atmospheric air is composed of

1) A mixture of gases (Nitrogen, Oxygen, Argon and Carbon dioxide etc.)

* Remarkably constant up to 100 km height (except for CO2);

* These four gases are the main components of dry air.

The specific gas constant: The mean molecular weight:

2) Water substance in any of its three physical states (vapor, droplets and ice particles)

* very important in radiative processes, cloud formation and interaction with the oceans, and highly variable.

3) Solid or liquid particles of very small size (atmospheric aerosols)

1105.287 KkgJRd196.28 molgM d

Problem: Find the average molecular weight M and specific constant R for air saturated with water vapor at 0oC and 1 atm of total pressure. The vapor pressure of water at 0oC is 6.11mb.

2.9 Hydrostatic balance

When an incremental air column experiences no net force in the vertical direction, it is considered to be in hydrostatic balance (or hydrostatic equilibrium).

0)( gdAdzdAdpppdA

is the acceleration of gravity.g

From

we have

gdz

dp (16)

Homework (1)

1. Using the equations of state for the ith gas and the mixture of gases, demonstrate that

.V

V

p

p

n

nN iiii

3. Problem 3.4 d) and e)

4. a) Determine the mean molar mass of the atmosphere of Venus, which consists of 95% CO2

and 5% N2 by volume.b) What is the corresponding gas constant?c) The mean surface temperature T on Venus is a scorching 740K as compared to only 288Kfor Earth; the surface pressure is 90 times that on Earth. By what factor is the density of the near-surface Venusian atmosphere greater or less than that of Earth?

5. Two sealed containers with volumes V1 and V2, respectively, contain dry air at pressures p1 and p2 and room temperature T. The containers are connected by a thin tube (negligible volume)That can be opened with a valve. When the valve is opened, the pressures equalize, and the system reequilibrates to room temperature. Find an expression for the new pressure.

6. Show that 1 atm of pressure is equivalent to that exerted by a 760 mm column of mercury at 0oC (density is 13.5951 g cm-3) and standard gravity g =9.8 ms-2.

2. Test the following equations for exactness. If it is exact, find the point function.

dyxydxyxz )()6(